2. Centripetal Acceleration
3. Negative Acceleration.
4. Positive Acceleration.
Use the Quizlet provided below:
https://quizlet.com/223675738/unit-1-lesson-1-lesson-2-review-flash-cards/
The temperature scale which starts at absolute zero is the Kelvin scale. The correct option in respect to the given question is the last option. William Thompson was the British scientist and inventor that invented the Kelvin scale. William Thompson was also popularly known as Lord Kelvin.His discovery of the Kelvin scale is considered one among the three best scales in use for measuring temperatures.Each measuring unit of this scale is never called a degree but a Kelvin. This specialized scale gives the option of measuring temperature in both centigrade and Fahrenheit.
Answer:
the angle is about 67.79 degrees
Explanation:
We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)
We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :
![half\,\,max-height = \frac{v_{yi}^2}{4\,g}](https://tex.z-dn.net/?f=half%5C%2C%5C%2Cmax-height%20%3D%20%5Cfrac%7Bv_%7Byi%7D%5E2%7D%7B4%5C%2Cg%7D)
we can use this information to find the y component of the velocity at that height via the formula:
![v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}](https://tex.z-dn.net/?f=v_%7Byf%7D%5E2-v_%7Byi%7D%5E2%3D-2%5C%2Cg%5C%2C%5CDelta%20y%5C%5C%5C%5Cv_%7Byf%7D%5E2-v_%7Byi%7D%5E2%3D-2%5C%2Cg%5C%2C%28%5Cfrac%7Bv_%7Byi%7D%5E2%7D%7B4%5C%2Cg%7D%20%29%5C%5Cv_%7Byf%7D%5E2%3Dv_%7Byi%7D%5E2-%5Cfrac%7Bv_%7Byi%7D%5E2%7D%7B2%7D%20%5C%5Cv_%7Byf%7D%5E2%3D%5Cfrac%7Bv_%7Byi%7D%5E2%7D%7B2%7D)
Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:
![1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\](https://tex.z-dn.net/?f=1%3D%5Csqrt%7Bv_%7Byf%7D%5E2%2B0.5%5E2%7D%20%5C%5C1%3D%5Csqrt%7B%5Cfrac%7Bv_%7Byi%7D%5E2%7D%7B2%7D%20%2B0.5%5E2%7D%5C%5C1%5E2%3D%5Cfrac%7Bv_%7Byi%7D%5E2%7D%7B2%7D%20%2B0.5%5E2%7D%5C%5C1-0.5%5E2%3D%5Cfrac%7Bv_%7Byi%7D%5E2%7D%7B2%7D%20%5C%5C2%281-0.5%5E2%29%3Dv_%7Byi%7D%5E2%5C%5C1.5%20%3D%20v_%7Byi%7D%5E2%5C%5Cv_%7Byi%7D%3D%5Csqrt%7B1.5%7D%20%5C%5C)
Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:
![tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o](https://tex.z-dn.net/?f=tan%28%5Ctheta%29%3D%5Cfrac%7Bv_%7Byi%7D%7D%7Bv_%7Bxi%7D%7D%20%3D%5Cfrac%7B%5Csqrt%7B1.5%7D%20%7D%7B0.5%7D%20%5C%5C%5Ctheta%3D%20arctan%28%5Cfrac%7B%5Csqrt%7B1.5%7D%20%7D%7B0.5%7D%29%3D67.79%5Eo)
I believe the answer is true